A force is represented by `F=ax^(2)+bt^((1)/(2))` where `x=` distance and `t=` time. The dimensions of `b^(2)//a` are:

Options

A. `[ML^(2)T^(-3)]`
B. `[ML^(-1)T^(-1)]`
C. `[MLT^(2)]`
D. `[ML^(3)T^(-3)]`

Answer

To solve the problem, we need to find the dimensions of \( \frac{b^2}{a} \) given the force equation \( F = ax^2 + bt^{1/2} \).

### Step-by-Step Solution

1. **Identify the Dimensions of Force**:
The dimension of force \( F \) is given by:
\[ [F] = M L T^{-2} \]

2. **Write the Expression for Force**:
The force is expressed as:
\[ F = ax^2 + bt^{1/2} \]

3. **Determine the Dimensions of Each Term**:
- For the term \( ax^2 \):
- The dimension of \( x \) (distance) is \( [x] = L \).
- Therefore, \( [ax^2] = [a][x^2] = [a][L^2] \).
- For the term \( bt^{1/2} \):
- The dimension of \( t \) (time) is \( [t] = T \).
- Therefore, \( [bt^{1/2}] = [b][t^{1/2}] = [b][T^{1/2}] \).

4. **Equate Dimensions**:
Since both terms on the right side must have the same dimension as \( F \), we can write:
\[ [a][L^2] = M L T^{-2} \quad \text{(1)} \]
\[ [b][T^{1/2}] = M L T^{-2} \quad \text{(2)} \]

5. **Solve for Dimensions of \( a \)**:
From equation (1):
\[ [a] = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

6. **Solve for Dimensions of \( b \)**:
From equation (2):
\[ [b] = \frac{M L T^{-2}}{T^{1/2}} = M L T^{-5/2} \]

7. **Calculate \( \frac{b^2}{a} \)**:
Now we need to find the dimensions of \( \frac{b^2}{a} \):
\[ [b^2] = (M L T^{-5/2})^2 = M^2 L^2 T^{-5} \]
\[ [a] = M L^{-1} T^{-2} \]
Therefore,
\[ \frac{b^2}{a} = \frac{M^2 L^2 T^{-5}}{M L^{-1} T^{-2}} = \frac{M^2}{M} \cdot \frac{L^2}{L^{-1}} \cdot \frac{T^{-5}}{T^{-2}} = M^{1} L^{3} T^{-3} \]

8. **Final Result**:
Thus, the dimensions of \( \frac{b^2}{a} \) are:
\[ [\frac{b^2}{a}] = M L^3 T^{-3} \]